Haskell 中的函数没有获取泛型类型参数

Question

In Haskell I have the following

data Box a = Box a deriving Show

instance Functor Box where
  fmap func (Box a) = Box (func a)

box :: Box Int
box = Box 8

wrapped :: Box (Box Int)
wrapped = Box <$> box

unwrap :: (Box a) -> a
unwrap (Box val) = val

In GHCi I can call fmap with unwrap wrapped and I get Box 8 , which I would expected.

*Main> fmap unwrap wrapped
Box 8

When I call fmap with unwrap box I get error about the types not matching, but I was thinking I would get the value 8.

*Main> fmap unwrap box

<interactive>:26:13: error:
    • Couldn't match type ‘Int’ with ‘Box b’
      Expected type: Box (Box b)
        Actual type: Box Int
    • In the second argument of ‘fmap’, namely ‘box’
      In the expression: fmap unwrap box
      In an equation for ‘it’: it = fmap unwrap box
    • Relevant bindings include
        it :: Box b (bound at <interactive>:26:1)

I would have expected fmap unwrap box would give the value 8 .

How would I define unwrap to be able to get the value of Box 8 for wrapped and 8 for box when using fmap ?

I do not think it matters but I am using GHCi version 8.8.4

Answer 1

If you have a box = Box 8 and want to get a plain 8 , you should just call unwrap box , not fmap unwrap box . In general, fmap will never let you unwrap a value because its type (Functor f) => (a -> b) -> fa -> fb specifies that the result of fmap xy will have the wrapped type fb . Let's walk through the types to see why this happens:

When we call fmap unwrap , we can "unify" the type of unwrap with the type of the first argument to fmap , which gives us this specialized type for fmap :

fmap   :: (Functor f) => (a     -> b) -> f a       -> f b
unwrap ::                 Box a -> a
fmap   :: (Functor f) => (Box a -> a) -> f (Box a) -> f a

Notice that because f doesn't appear in the type of the first argument of fmap , Box doesn't fill in for f . We can then apply unwrap to this specialized version of fmap :

fmap unwrap :: (Functor f) => f (Box a) -> f a

This is almost certainly not the type you're expecting here. When we unify this for the argument Box (Box 8) , we get the specialized type:

fmap unwrap :: (Functor f) => f   (Box a  ) -> f   a
Box (Box 8) ::                Box (Box Int)
fmap unwrap ::                Box (Box Int) -> Box Int

But when we try to unify this with the type of argument Box 8 , we run into a problem:

fmap unwrap :: (Functor f) => f   (Box a) -> f   a
Box 8       ::                Box Int

I can't match the type of Box 8 with the expected argument to fmap unwrap because Int doesn't match the type Box a .